To find constant term of polynomial defined recursively 
Let $\{f_n(x)\}$ be a sequence of polynomials defined inductively as
  \begin{align*}
     f_1(x) &= (x-2)^2, \\
 f_{n+1}(x) &= (f_n(x)-2)^2, \quad n \geq 1.
\end{align*}
  Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f_n(x)$. Then
  
  
*
  
*$a_n = 4$, $b_n = -4^n$,
  
*$a_n = 4$, $b_n = -4n^2$,
  
*$a_n = 4^{(n-1)!}$, $b_n = -4^n$,
  
*$a_n = 4^{(n-1)!}, b_n = -4n^2$.
  

I see that general or $n$-th term of this will be $((x-2)^2 - 2)^2-2)^2...$ But how do I simplify this to get to the constant term 
 A: The trick is to ignore $x^2$ terms because they will never affect $a_n$ or $b_n$ in later values.
$f_1(x)=x^2g_1(x)-4x+4$
$f_2(x)=...+(-4x+2)^2=...-16x+4$
$f_3(x)=...+(-16x+2)^2=...-64x+4$
$f_4(x)=...+(-64x+2)^2=...-256x+4$
A: To expand on the comment by Vinod, for any polynomial, we get the constant term by plugging 0 in for $x$. In fact, that's pretty much the definition of the "constant term". Doing this we see that we will get
$$\begin{align*}
(\cdots (0-2)^2 - 2)^2 - \cdots - 2)^2 &= (\cdots ((-2)^2-2)^2 - 2)^2 - \cdots - )^2 \\
&= (\cdots (4-2)^2 - 2)^2 - \cdots - 2)^2 \\
&= (\cdots (2)^2 - 2)^2 - \cdots - 2)^2 \\
&= (\cdots (4 - 2)^2 - \cdots - 2)^2 \\
&= \cdots \\
&= 4
\end{align*}$$
(We continually remove a set of parentheses in each step, and eventually end up with $4$ as our constant term.)
Now onto the coefficient of $x$, we can perform a similar analysis. We first note that the coefficient of $x$ for $n=1$ is $-4$:
$$f_1(x) = (x-2)^2 = x^2 - 4x + 4$$
Now we prove by induction that the coefficient is $-4^n$.
Assume that the coefficient of $x$ is $-4^n$ for $f_n$. Then we know $f_{n+1}(x)$ is defined as
$$ f_{n+1}(x) = (f_n(x) - 2)^2$$
We know three very important things about this equation. First, by the inductive hypothesis, we know that the coefficient of $x$ for $f_n$ is $-4^n$, we know by the previous step that the constant term of $f_n$ is $4$, and we know that we don't care about the other terms. Observe:
$$\begin{align*}
(f_n(x) - 2)^2 &= (c_n x^n + \cdots + c_2 x^2 - 4^n x + 4 - 2)^2 \\
&= (c_n x^n + \cdots + c_2 x^2 - 4^n x + 2)^2 \\
&= (c_n x^n + \cdots + c_2 x^2 - 4^n x + 2) \cdot (c_n x^n + \cdots + c_2 x^2 - 4^n x + 2) \\
&= \cdots - 4^n x \cdot 2 - 4^n x \cdot 2 + 2 \cdot 2 \\
&= \cdots - 4 \cdot 4^n x + 4 \\
&= \cdots - 4^{n+1} x + 4
\end{align*}$$
as desired. Note that the $\cdots$ in the last few equations are all higher order terms, i.e. when multiplying $(c_n x^n + \cdots + c_2 x^2 - 4^n x + 2)$ by itself, the $\cdots$ represent all of the terms whose exponents are 2 or greater.
Finally, it is worth noting that once you know the constant term is 4, you can manually calculate the $f_1$, $f_2$, and $f_3$ by hand and see that the coefficients for $x$ are $-4^1 = -4$, $-4^2 = -16$, and $-4^3 = -64$ respectively, and then conclude (since it is a multiple choice problem), that the answer is $a_n = 4$ and $b_n = -4^n$, but where's the fun in that?
